Question

Find all points of discontinuity of f, where f is defined by

$f(x) = \begin{cases} x^{10}-1, & \quad \text{if } x{\leq 1}\\\ x^2, & \quad \text{if } x \text{>1} \end{cases}$

Answer 1

$f(x) = \begin{cases} x^{10}-1, & \quad \text{if } x{\leq 1}\\\ x^2, & \quad \text{if } x \text{>1} \end{cases}$
The given function f is defined at all the points of the real line.
Let c be a point on the real line.

Case (i):
If c<1, then f(c) = c10-1 and $\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ f(x10-1) = c10-1
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x<1

Case (ii):
If c = 1,then the left hand limit of f at x = 1 is
$\lim\limits_{x \to 1^-}$ f(x) =$\lim\limits_{x \to 1^-}$(x10-1)=1-1=0
The right hand limit of f at x = 1 is,
$\lim\limits_{x \to 1^+}$ f(x) = $\lim\limits_{x \to 1^+}$(x2) = 12 = 1
It is observed that the left and right hand limit of f at x = 1 do not coincide.
Therefore,f is not continuous at x = 1

Case(iii):
Ifc>1, then f(c) = c2
$\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ (x2) = c2
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x>1

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.